HELLPP!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

6 0

$\left\{\begin{array}{ccc}2x-3y=-27&|\cdot3\\-3x+2y=23&|\cdot2\end{array}\right\\\underline{\left\{\begin{array}{ccc}6x-9y=-81\\-6x+4y=46\end{array}\right}\ \ \ \ |\text{add both sides}\\.\ \ \ \ \ \ -5y=-35\ \ \ \ \ |:(-5)\\.\ \ \ \ \ \ \ \ y=7\\\\\text{put the value of y to the first equation}\\\\2x-3(7)=-27\\\\2x-21=-27\ \ \ \ |+21\\\\2x=-6\ \ \ \ |:2\\\\x=-3\\\\\text{Answer:}\ x=-3,\ y=7\to\boxed{(-3,\ 7)}$

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What is ∛1728. Then multiplied by ∛14903

NeTakaya

Cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.

Solution:

Need to calculate $\sqrt[3]{1728}$ and then multiply the result by $\sqrt[3]{14903}$

Let us first evaluate $\sqrt[3]{1728}$

$\Rightarrow \sqrt[3]{1728}=\sqrt[3]{12 \times 12 \times 12}=12$

As need to multiply 12 by $\sqrt[3]{14903}$

$\Rightarrow 12 \times \sqrt[3]{14903}$

On solving $\sqrt[3]{14903}$ , we get

$\sqrt[3]{14903}=24.608$

$\Rightarrow 12 \times \sqrt[3]{14903}=12 \times 24.608=295.306$

Hence cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.

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